Sure! Let's break down the analysis of the function [tex]\( f(x) = \log_2(x) \)[/tex] according to the points listed:
1. x-intercept of [tex]\( f(x) \)[/tex]:
- The [tex]\( x \)[/tex]-intercept is the point where the function [tex]\( f(x) \)[/tex] crosses the [tex]\( x \)[/tex]-axis. This occurs when [tex]\( f(x) = 0 \)[/tex].
- Setting [tex]\( f(x) \)[/tex] to zero: [tex]\( \log_2(x) = 0 \)[/tex].
- Solving for [tex]\( x \)[/tex]: [tex]\( x = 2^0 = 1 \)[/tex].
- Therefore, the [tex]\( x \)[/tex]-intercept of [tex]\( f(x) = \log_2(x) \)[/tex] is at [tex]\( x = 1 \)[/tex].
2. Vertical asymptote of [tex]\( f(x) \)[/tex]:
- A vertical asymptote occurs where the function approaches infinity as [tex]\( x \)[/tex] approaches a certain value.
- For [tex]\( \log_2(x) \)[/tex], the function is not defined for [tex]\( x \leq 0 \)[/tex], and as [tex]\( x \)[/tex] approaches 0 from the positive side, [tex]\( \log_2(x) \)[/tex] tends to -∞.
- Therefore, the vertical asymptote of [tex]\( f(x) = \log_2(x) \)[/tex] is at [tex]\( x = 0 \)[/tex].
3. Intersection of [tex]\( f(x) \)[/tex] and the line [tex]\( y = 1 \)[/tex]:
- To find where the function [tex]\( f(x) \)[/tex] intersects the line [tex]\( y = 1 \)[/tex], set [tex]\( f(x) \)[/tex] equal to 1.
- Setting [tex]\( \log_2(x) = 1 \)[/tex].
- Solving for [tex]\( x \)[/tex]: [tex]\( x = 2^1 = 2 \)[/tex].
- Therefore, the intersection of [tex]\( f(x) = \log_2(x) \)[/tex] and the line [tex]\( y = 1 \)[/tex] is at [tex]\( x = 2 \)[/tex].
Based on these points, Otto should conclude the following about the function [tex]\( f(x) = \log_2(x) \)[/tex]:
- The [tex]\( x \)[/tex]-intercept is at [tex]\( x = 1 \)[/tex].
- There is a vertical asymptote at [tex]\( x = 0 \)[/tex].
- The function intersects the line [tex]\( y = 1 \)[/tex] at [tex]\( x = 2 \)[/tex].
Thus, the correct answer Otto should record is:
[tex]\[ (1, 0, 2) \][/tex]
